Leetcode: Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             / \            4   8           /   / \          11  13  4         /  \    / \        7    2  5   1

return

[   [5,4,11,2],   [5,8,4,5]]

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {// Start typing your Java solution below// DO NOT write main() functionArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();ArrayList<Integer> path = new ArrayList<Integer>();traversal(res, path, root, sum);return res;}public void traversal(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> path, TreeNode root, int sum){if(root == null)return;path.add(root.val);sum = sum - root.val;if(root.left == null && root.right == null && sum == 0)res.add(new ArrayList<Integer>(path));traversal(res, path, root.left, sum);traversal(res, path, root.right, sum);path.remove(path.size() - 1);}}

555...代码一遍过，不容易啊

发现其实还可以剪枝，如果过程中sum已经小于零了，直接这条路径不再继续return